If sin q and cos q are the roots of the equation ax2 +bx + c = 0, then

Since sin q and cos q are the roots of the equation ax2 +bx + c = 0∴ sin⁡θ+cos⁡θ=−baandsin⁡θcos⁡θ=caNow (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ∴b2a2=1+2ca=a+2ca∴ b2=a(a+2c)=a2+2ac⇒b2+c2=a2+2ac+c2=(a+c)2⇒(a+c)2=b2+c2

If sin q and cos q are the roots of the equation ax2 +bx + c = 0, then

Since sin q and cos q are the roots of the equation ax2 +bx + c = 0∴ sin⁡θ+cos⁡θ=−baandsin⁡θcos⁡θ=caNow (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ∴b2a2=1+2ca=a+2ca∴ b2=a(a+2c)=a2+2ac⇒b2+c2=a2+2ac+c2=(a+c)2⇒(a+c)2=b2+c2

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If sin q and cos q are the roots of the equation ax² +bx + c = 0, then

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$(a - c)^{2} = b^{2} - c^{2}$

$(a + c)^{2} = b^{2} - c^{2}$

$(a + c)^{2} = b^{2} + c^{2}$

$(a - c)^{2} = b^{2} + c^{2}$

answer is D.

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Detailed Solution

Since sin q and cos q are the roots of the equation ax² +bx + c = 0

$\begin{array}{ll} ∴ & \sin θ + \cos θ = - \frac{b}{a} and \sin θcos θ = \frac{c}{a} \\ Now (\sin θ + \cos θ)^{2} = 1 + 2 \sin θcos θ & ∴ \frac{b^{2}}{a^{2}} = 1 + \frac{2 c}{a} = \frac{a + 2 c}{a} \\ ∴ & b^{2} = a (a + 2 c) = a^{2} + 2 ac \\ \Rightarrow & b^{2} + c^{2} = a^{2} + 2 ac + c^{2} = (a + c)^{2} \\ \Rightarrow & (a + c)^{2} = b^{2} + c^{2} \end{array}$